Colebrook friction factor 4

[Zapster]

What are the origins for different Colebrook friction factor equations shown in a few texts?

This one I have used in the past is 1/f^(-0.5) = -2Log(?/(3.7Re)+2.51/(f^0.5*Re))

Not sure why John McKetta and Ron Darby use these coefficients: 1/f^(-0.5) = -4Log(?/(3.7Re)+1.26/(f^0.5*Re))

 

[katmar]

There are two main forms of the friction factor. One is known as the Fanning friction factor and the other the Moody friction factor (Moody factor is also known as Darcy factor). Moody is simply 4x the Fanning factor. In your two equations the coefficients vary by 2x because of this 4x factor and the square root on the left hand side.

Have a look at the FAQs at the top of this forum - some references are given in the FAQ written by Quark.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[BigInch]

Zapster,
I know its not the question you asked, but you can save yourself a lot of grief if you switch to the Churchill equation.

http://www.eng-tips.com/faqs.cfm?fid=1236

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[katmar]

I agree with BigInch that the Churchill equation is very useful, but introducing it into the mix requires a word of caution.

In my earlier post I said that there are two main forms of the Colebrook friction factor - the Fanning and Moody (or Darcy) factors. There is in fact a third form, which I call the Stanton friction factor. This friction factor is 1/2 of the Fanning factor and 1/8 of the Moody factor. It is very important to know which friction factor you are using because the coefficient in the Darcy-Weisbach equation for frictional pressure drop must change accordingly.

The Stanton form is very rarely used, but one of the authors who does use it is Churchill. The original publication by Churchill (Chem Eng, Nov 7, 1977) and the form given by Quark in the FAQ above both use the Stanton friction factor.

Many young (and not so young) engineers have discovered that their pressure drop calculations are out by a factor of 2, 4 or 8 depending on which friction factor they have used with which form of the D-W equation.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Zapster]

katmar and BigInch, thank you for the helpful replies.